# A balanced lever has two weights on it, the first with mass 3 kg  and the second with mass 4 kg. If the first weight is  8 m from the fulcrum, how far is the second weight from the fulcrum?

Jan 23, 2017

6 m away from the fulcrum, opposite from the first weight.

#### Explanation:

For the lever to be "balanced", there must be no net torque on the system, or

${\tau}_{\text{net}} = 0$

or

$\Sigma \tau = 0$

Assuming the fulcrum is at the center of mass of the lever (or the lever is massless), the torques created by each of the weights must be equal and opposite:

${\tau}_{1} + {\tau}_{2} = 0$
${\tau}_{1} = - {\tau}_{2}$

Torque is the force perpendicular to a point of rotation multiplied by the distance from that pivot point. Assuming the lever to be horizontal, the torque equation becomes:

${F}_{1} \times {r}_{1} = - \left({F}_{2} \times {r}_{2}\right)$
${m}_{1} g \times {r}_{1} = - \left({m}_{2} g \times {r}_{2}\right)$

Solving for ${r}_{2}$ gives:

${r}_{2} = \frac{\left({m}_{1} \cancel{g}\right) {r}_{1}}{- {m}_{2} \cancel{g}} = - {m}_{1} / {m}_{2} {r}_{1}$

Applying the given information, we get:

r_2=-(3"kg")/(4"kg")(8"m")=-6"m"

The fact that ${r}_{2}$ is a negative value tells us that it is on the opposite side of the fulcrum from the first weight.

The second weight must be positioned 6 m away from the fulcrum on the other side of the fulcrum.